The uniform distribution

In Example: Uncertainties in steel ball manufacturing, we had a random variable for which all values between a given interval were equally probable. This is a very common situation covered by the so-called uniform distribution.

Let’s start with the simplest case: a random variable taking values between 0 and 1 with constant probability density. We write:

\[ X\sim U([0,1]), \]

and we read

\(X\) follows a uniform distribution taking values in \([0,1]\).

The probability density of the uniform is constant in \([0,1]\) and zero outside it. We have:

\[\begin{split} p(x) := \begin{cases} c,&\;0\le x \le 1,\\ 0,&\;\text{otherwise}. \end{cases} \end{split}\]

What should the constant \(c\) be? Just like in Example: Uncertainties in steel ball manufacturing, you can find it by ensuring the PDF integrates to one (see PDF Property 5:

\[ \int_0^1 p(x) dx = 1 \Rightarrow c = 1. \]

So, the PDF is:

\[\begin{split} p(x) := \begin{cases} c,&\;0\le x \le 1,\\ 0,&\;\text{otherwise}. \end{cases} \end{split}\]

To find the CDF, we can use PDF Property 3: $\( F(x) = p(X \le x) = \int_0^x p(\tilde{x}) d\tilde{x} = \int_0^x d\tilde{x} = x. \)\( Obviously, we have \)F(x) = 0\( for \)x < 0\( and \)F(x) = 1\( for \)x > 1$.

Using this result, we can find the probability that \(X\) takes values in \([a,b]\) for \(a < b\) in \([0,1]\). It is: $\( p(a \le X \le b) = F(b) - F(a) = b - a. \)$

Instantiating the uniform using scipy.stats

Let me know show you how you can make a uniform random variable using scipy:

Show code cell source Hide code cell source

import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
sns.set(rc={"figure.dpi":100, 'savefig.dpi':300})
sns.set_context('notebook')
sns.set_style("ticks")
from IPython.display import set_matplotlib_formats
set_matplotlib_formats('retina', 'svg')
import numpy as np
import scipy.stats as st

X = st.uniform()

Here is how you can get some samples:

X.rvs(size=10)

array([0.33842053, 0.57010738, 0.1323871 , 0.58357996, 0.68875828,
       0.67923023, 0.84656861, 0.75593012, 0.25791332, 0.13214875])

You can evalute the PDF at any point like this:

X.pdf(0.5)

1.0

X.pdf(-0.1)

0.0

X.pdf(1.5)

0.0

Let’s plot the PDF:

fig, ax = plt.subplots()
xs = np.linspace(-0.1, 1.1, 200)
ax.plot(xs, X.pdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$');

You can evaluate the CDF like this:

X.cdf(-0.5)

0.0

X.cdf(0.5)

0.5

X.cdf(1.2)

1.0

Let’s plot the CDF:

fig, ax = plt.subplots()
ax.plot(xs, X.cdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$F(x)$');

Finally, let’s find the probability that \(X\) is between two numbers \(a\) and \(b\). For the uniform it is, of course, trivial, but let’s see how it is done using the scipy functionality:

a = -1.0
b = 0.3
prob_X_is_in_ab = X.cdf(b) - X.cdf(a)
print('p({0:1.2f} <= X <= {1:1.2f}) = {2:1.2f}'.format(a, b, prob_X_is_in_ab))

p(-1.00 <= X <= 0.30) = 0.30

The uniform distribution over an arbitrary interval \([a, b]\)

The uniform distribution can also be defined over an arbitrary interval \([a,b]\). We write: $\( X \sim U([a, b]). \)$ We read:

\(X\) follows a uniform distribution on \([a, b]\).

The PDF of this random variable is:

\[\begin{split} p(x) = \begin{cases} c,&\;x\in[a,b],\\ 0,&\;\text{otherwise}, \end{cases} \end{split}\]

where \(c\) is a positive constant. This simply tells us that the probability density of finding \(X\) in \([a,b]\) is something positive and that the probability density of findinig outside is exactly zero. This is exactly the situation we had in Example: Uncertainties in steel ball manufacturing. The positive constant \(c\) is determined by imposing the normalization condition:

\[ \int_{-\infty}^{+\infty}p(x)dx = 1. \]

This gives:

\[ 1 = \int_{-\infty}^{+\infty}p(x)dx = \int_a^bc dx = c \int_a^bdx = c (b-a). \]

From this we get:

\[ c = \frac{1}{b - a}, \]

and we can now write:

\[\begin{split} p(x) = \begin{cases} \frac{1}{b-a},&x \in [a, b],\\ 0,&\;\text{otherwise}, \end{cases} \end{split}\]

From the PDF, we can now find the CDF for \(x \in [a,b]\):

\[ F(x) = p(X\le x) = \int_{-\infty}^x p(\tilde{x})d\tilde{x} = \int_a^x \frac{1}{b-a}d\tilde{x} = \frac{1}{b-a}\int_a^xd\tilde{x} = \frac{x-a}{b-a}. \]

Instantiating the generic uniform using scipy.stats:

Let’s instantiate using scipy.stats:

a = -2.0
b = 5.0
X = st.uniform(loc=a, scale=(b-a))

Some samples:

X.rvs(size=10)

array([-1.63839392, -1.08615066, -1.9249188 , -0.86574048, -0.01377033,
        0.91441   ,  2.94486672, -0.90602192, -0.44926847,  4.28172985])

The PDF:

fig, ax = plt.subplots()
xs = np.linspace(a - 0.2, b + 0.2, 200)
ax.plot(xs, X.pdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$');

The CDF:

fig, ax = plt.subplots()
ax.plot(xs, X.cdf(xs))
ax.set_xlabel('$x$')
ax.set_ylabel('$F(x)$');

Questions

• Repeat the code above so that the random variable is \(U([1, 10])\).